46) Determine solutions for each equation in the interval 0 ≤ x ≤ 2π, to the nearest hundredth of a radian.
Give exact answers where possible.
h) 2 csc2 x − 9 csc x + 10 = 0
To solve the equation 2 csc^2(x) - 9 csc(x) + 10 = 0, we can substitute u = csc(x). This allows us to rewrite the equation as 2u^2 - 9u + 10 = 0.
Now, we can solve this quadratic equation for u:
2u^2 - 9u + 10 = 0
Using the quadratic formula, we have:
u = (-(-9) ± √((-9)^2 - 4(2)(10))) / (2(2))
= (9 ± √(81 - 80)) / 4
= (9 ± √1) / 4
We get two solutions for u:
u1 = (9 + 1) / 4
= 10 / 4
= 5/2
u2 = (9 - 1) / 4
= 8 / 4
= 2
Now that we have the values of u, we can find the values of x by taking the inverse csc function of u.
For u1 = 5/2:
x = csc^(-1)(5/2)
Using a calculator, we find:
x ≈ 0.5236 radians (rounded to the nearest hundredth)
For u2 = 2:
x = csc^(-1)(2)
Using a calculator, we find:
x ≈ 1.2309 radians (rounded to the nearest hundredth)
Therefore, the solutions to the equation 2 csc^2(x) - 9 csc(x) + 10 = 0 in the interval 0 ≤ x ≤ 2π, to the nearest hundredth of a radian, are:
x ≈ 0.5236 radians and x ≈ 1.2309 radians.